Question
Evaluate the following integrals:
$\int\text{x}^3\tan^{-1}\text{x dx}$
$\int\text{x}^3\tan^{-1}\text{x dx}$
✓
Answer
$\int\text{x}^3\tan^{-1}\text{x dx}$
$\int\text{x}^3\tan^{-1}\text{x dx}=\tan^{-1}\text{x}\int\text{x}^3\text{dx}-\Big(\int\frac{\text{d}\tan^{-1}\text{x}}{\text{dx}}\big(\int\text{x}^3\text{dx}\big)\text{dx}\Big)$
$=\tan^{-1}\text{x}\frac{\text{x}^4}{4}-\Big(\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}\Big)$
$=\tan^{-1}\text{x}\frac{\text{x}^4}{4}-\Big(\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}\Big)$
$\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}=\frac{1}{4}\Big[\int\frac{1}{1+\text{x}^2}\text{dx}+(\text{x}^2-1)\text{dx}\Big]$
$\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}=\frac{1}{4}\Big[\tan^{-1}\text{x}+\frac{\text{x}^3}{3}-\text{x}\Big]$
$\int\text{x}^3\tan^{-1}\text{x dx}=\frac{\text{x}^4}{4}\tan^{-1}\text{x}-\frac{1}{4}\Big[\tan^{-1}\text{x}+\frac{\text{x}^3}{3}-\text{x}\Big]+\text{C}$
$\int\text{x}^3\tan^{-1}\text{x dx}=\tan^{-1}\text{x}\int\text{x}^3\text{dx}-\Big(\int\frac{\text{d}\tan^{-1}\text{x}}{\text{dx}}\big(\int\text{x}^3\text{dx}\big)\text{dx}\Big)$
$=\tan^{-1}\text{x}\frac{\text{x}^4}{4}-\Big(\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}\Big)$
$=\tan^{-1}\text{x}\frac{\text{x}^4}{4}-\Big(\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}\Big)$
$\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}=\frac{1}{4}\Big[\int\frac{1}{1+\text{x}^2}\text{dx}+(\text{x}^2-1)\text{dx}\Big]$
$\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}=\frac{1}{4}\Big[\tan^{-1}\text{x}+\frac{\text{x}^3}{3}-\text{x}\Big]$
$\int\text{x}^3\tan^{-1}\text{x dx}=\frac{\text{x}^4}{4}\tan^{-1}\text{x}-\frac{1}{4}\Big[\tan^{-1}\text{x}+\frac{\text{x}^3}{3}-\text{x}\Big]+\text{C}$
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